Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
| 1: |
|
minus(X,s(Y)) |
→ pred(minus(X,Y)) |
| 2: |
|
minus(X,0) |
→ X |
| 3: |
|
pred(s(X)) |
→ X |
| 4: |
|
le(s(X),s(Y)) |
→ le(X,Y) |
| 5: |
|
le(s(X),0) |
→ false |
| 6: |
|
le(0,Y) |
→ true |
| 7: |
|
gcd(0,Y) |
→ 0 |
| 8: |
|
gcd(s(X),0) |
→ s(X) |
| 9: |
|
gcd(s(X),s(Y)) |
→ if(le(Y,X),s(X),s(Y)) |
| 10: |
|
if(true,s(X),s(Y)) |
→ gcd(minus(X,Y),s(Y)) |
| 11: |
|
if(false,s(X),s(Y)) |
→ gcd(minus(Y,X),s(X)) |
|
There are 9 dependency pairs:
|
| 12: |
|
MINUS(X,s(Y)) |
→ PRED(minus(X,Y)) |
| 13: |
|
MINUS(X,s(Y)) |
→ MINUS(X,Y) |
| 14: |
|
LE(s(X),s(Y)) |
→ LE(X,Y) |
| 15: |
|
GCD(s(X),s(Y)) |
→ IF(le(Y,X),s(X),s(Y)) |
| 16: |
|
GCD(s(X),s(Y)) |
→ LE(Y,X) |
| 17: |
|
IF(true,s(X),s(Y)) |
→ GCD(minus(X,Y),s(Y)) |
| 18: |
|
IF(true,s(X),s(Y)) |
→ MINUS(X,Y) |
| 19: |
|
IF(false,s(X),s(Y)) |
→ GCD(minus(Y,X),s(X)) |
| 20: |
|
IF(false,s(X),s(Y)) |
→ MINUS(Y,X) |
|
The approximated dependency graph contains 3 SCCs:
{14},
{13}
and {15,17,19}.
-
Consider the SCC {14}.
There are no usable rules.
By taking the AF π with
π(LE) = 1 together with
the lexicographic path order with
empty precedence,
rule 14
is strictly decreasing.
-
Consider the SCC {13}.
There are no usable rules.
By taking the AF π with
π(MINUS) = 2 together with
the lexicographic path order with
empty precedence,
rule 13
is strictly decreasing.
-
Consider the SCC {15,17,19}.
The usable rules are {1-6}.
The constraints could not be solved.
Tyrolean Termination Tool (0.26 seconds)
--- May 3, 2006